tolong di bantu yaa kak/mas.. gawat ni kalok gk siap gk bisa ujian :`(

(g¤f)(x)=(akar x+1)^2 -1
=x+1-1
=x
semoga menbantu:)
koreksi kalau salah ya

[tex]\displaystyle (g\circ f)(x)=g(f(x))\\(g\circ f)(x)=(f(x))^2-1\\(g\circ f)(x)=(\sqrt{x+1})^2-1\\(g\circ f)(x)=x+1-1\\\boxed{\boxed{(g\circ f)(x)=x}}[/tex]